Proof that f(z) is in no point analytic

[jennyjones]

Homework Statement

I have to proof for a homework assignment that f(z) = z^(n)(z*)^(-m) where n, m ≥ 1 (are natural numbers) is in no point analytic.

Homework Equations

binomial function

The Attempt at a Solution

I found this wikipedia page, were they state that:
a complex-valued function ƒ of a complex variable z is said to be analytic at a if in some open disk centered at a it can be expanded as a convergent power series.

for which i think i need to use the binomial function.

But I'm not sure how to do this for the given function f(z)

 

[jennyjones]

(z*) is the complex conjugate


If i look at the Lim (z->0) z/z* i would say that it does not exist because if i look let z approach 0 on the real axis i get 1, and if i let z approach 0 on the imaginary axis i get -1.

but now i think i have to look at the lim(z->0) z^n/(z*)^m and am not sure how to approach the powers.

thanx

 

[vela]

You might want to use a different criterion for analyticity for this problem.

http://en.wikipedia.org/wiki/Cauchy–Riemann_equations

I haven't worked out the problem, but that's where I would start.

 

[jennyjones]

So the function is analytic in a domain if it is differentiable at all points of the domain.

 

[Xiuh]

There is yet another criterion based on the so called Wirtinger derivatives

see: http://wcherry.math.unt.edu/math5410/wirtinger.pdf

Roughly, if $\frac{\partial f(z,\overline{z})}{\partial \overline{z}} \neq 0$, then $f$ does not satisfy the Cauchy-Riemann equations, and thus cannot be analytic.

Edit: Here $\overline{z}$ is the conjugate of $z$